Simplifying trig expressions

When you reach for it

Reach for these moves when an expression carries more than it needs: several different functions of the same angle, a fraction inside a fraction, two fractions waiting to be combined, or a square such as 1 - \sin^2 x that one identity collapses. A shorter form is easier to evaluate, graph and solve, which is why trigonometric equations usually start with a step like this.

The same identities also run the other way, into algebra. A radical such as \displaystyle \sqrt{9 - x^2} turns into a single trigonometric function when the variable is replaced by 3\sin\theta, because the identity 1 - \sin^2\theta = \cos^2\theta swallows the subtraction under the root. That replacement is called a trigonometric substitution.

To simplify is not to prove. Here the task is to start from one expression and find a shorter equal one. Showing that two given expressions are equal is verifying an identity, the next lesson.

The procedure

Everything rests on four families of identities, each from an earlier lesson.

• Reciprocal and quotient, holding wherever the left side exists: \displaystyle \sec x = \frac{1}{\cos x}, \displaystyle \csc x = \frac{1}{\sin x}, \displaystyle \tan x = \frac{\sin x}{\cos x}, \displaystyle \cot x = \frac{\cos x}{\sin x}.
• Pythagorean, holding wherever both sides exist: \sin^2 x + \cos^2 x = 1, 1 + \tan^2 x = \sec^2 x, 1 + \cot^2 x = \csc^2 x.
• Cofunction: \displaystyle \sin\left(\frac{\pi}{2} - x\right) = \cos x, \displaystyle \cos\left(\frac{\pi}{2} - x\right) = \sin x, \displaystyle \cot\left(\frac{\pi}{2} - x\right) = \tan x.
• Even and odd: \cos(-x) = \cos x and \sec(-x) = \sec x, while \sin(-x) = -\sin x, and likewise \tan, \csc and \cot are odd.

Rewrite in sine and cosine when the expression mixes functions. Each reciprocal and quotient identity holds wherever the function being replaced exists, so the swap never changes a value.

Look for a Pythagorean pair. The forms 1 - \sin^2 x = \cos^2 x, 1 - \cos^2 x = \sin^2 x, \sec^2 x - 1 = \tan^2 x and \csc^2 x - 1 = \cot^2 x are the ones that shorten expressions. A denominator 1 + \sin x or 1 - \cos x is half of such a pair, since (1 - \sin x)(1 + \sin x) = 1 - \sin^2 x.

Treat \sin x as a number. At each input, \sin x is a real number, so every law of algebra applies to it: factoring a quadratic in \sin x, combining fractions over a common denominator, the difference of squares. This is the algebra of Simplifying rational expressions with \sin x or \cos x in place of the variable.

Cancel only what is not zero, and keep the domain. Find the original expression’s domain before the first step, and quote the answer on it: “equals \tan x for \displaystyle x \neq \frac{k\pi}{2}”. Cancelling a factor can give a short form that exists at inputs the original excluded, which shows as holes in the original’s graph. Multiplying top and bottom by a new factor can do the reverse, and the new factor must be nonzero for the step to multiply by 1. Throughout, k is any integer.

There is no single simplest form: fewer functions, fewer terms and no fraction inside a fraction are the usual aims.

A variable replaced by a trigonometric function

To simplify \displaystyle \sqrt{9 - x^2}, set x = 3\sin\theta. Then 9 - x^2 = 9 - 9\sin^2\theta = 9\cos^2\theta, and the square root of a square is an absolute value (Simplifying radicals):

The bars come off only if the sign of \cos\theta is known, and that is what the interval is for. With \displaystyle -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}, the angle’s point lies on the right half of the circle (Figure 1), where \cos\theta \ge 0; so \displaystyle \sqrt{9 - x^2} = 3\cos\theta. The same interval lets 3\sin\theta take every value from -3 to 3 exactly once, so each x in the domain has one \theta, the one an inverse sine would return.

The three standard substitutions, for a constant a > 0:

• \displaystyle \sqrt{a^2 - x^2}: set x = a\sin\theta with \displaystyle -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}; the root becomes a\cos\theta.
• \displaystyle \sqrt{a^2 + x^2}: set x = a\tan\theta with \displaystyle -\frac{\pi}{2} < \theta < \frac{\pi}{2}; the root becomes a\sec\theta.
• \displaystyle \sqrt{x^2 - a^2} for x \ge a: set x = a\sec\theta with \displaystyle 0 \le \theta < \frac{\pi}{2}; the root becomes a\tan\theta.

In each row the interval is chosen so that the function left under the root is not negative there.

Worked examples

Boundaries

The short form stands in only on the original’s domain. Cancelling can make the short form exist at extra inputs (Examples 1, 2 and 4), which appear as holes in the original’s graph. Multiplying by a new factor can make it miss inputs where the original had a value (Example 5). Either way the result is an identity only on the stated set.

A cancelled factor must be nonzero. Dividing top and bottom by \sin x + 1 is allowed only where \sin x \neq -1, which is why that set leaves the domain.

The root of a square is an absolute value. \displaystyle \sqrt{1 - \sin^2 x} = |\cos x|, which equals \cos x only where \cos x \ge 0. A substitution drops the bars only because its interval fixes the sign; outside the interval the same algebra gives the negative.

A substitution covers only the values it reaches. x = 3\sin\theta reaches only -3 \le x \le 3, which is exactly where \displaystyle \sqrt{9 - x^2} exists. x = a\sec\theta with \displaystyle 0 \le \theta < \frac{\pi}{2} reaches only x \ge a; the values x \le -a need \displaystyle \frac{\pi}{2} < \theta \le \pi, where \tan\theta \le 0 and the root is -a\tan\theta.

Where this goes next

The same moves, aimed at a target, are how an identity is proved in Verifying identities: one side is simplified until it becomes the other. Simplifying first is also the opening step of most Solving trigonometric equations, where the domain bookkeeping decides which candidate solutions survive. The reverse of a substitution, getting from \theta back to an expression in x, is Composing trig and inverse trig functions, drawn with the same reference triangle as Figure 1. In calculus, trigonometric substitution is a standard way to handle \displaystyle \sqrt{a^2 - x^2}, \displaystyle \sqrt{a^2 + x^2} and \displaystyle \sqrt{x^2 - a^2}, in the same spirit as u-substitution.

Exercises

Rewrite and combine

Simplify, and state the set on which the result agrees with the original.

1. \cos x + \sin x\tan x
2. \displaystyle \frac{1 - \cos^2 x}{\sin x\cos x}
3. \csc x - \cos x\cot x
4. (\sec x - \tan x)(\sec x + \tan x)
5. \displaystyle \frac{\sin x}{\csc x} + \frac{\cos x}{\sec x}
6. \displaystyle \frac{1}{1 + \sin x} + \frac{1}{1 - \sin x}
7. \displaystyle \frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}

Even, odd and cofunction

Simplify, and state the set on which the result agrees with the original.

8. \sin(-x)\csc x + \cos^2(-x)
9. \displaystyle \sin\left(\frac{\pi}{2} - x\right)\tan x
10. \displaystyle \frac{\tan(-x)}{\cot\left(\frac{\pi}{2} - x\right)}

Factor first

11. Simplify \displaystyle \frac{\tan^2 x}{\sec x + 1}, state its domain, and find the holes in its graph.
12. Simplify \sin^4 x - \cos^4 x.
13. Simplify \displaystyle \frac{2\cos^2 x - \cos x - 1}{\sin^2 x} and find the holes in its graph.

Substitutions

Make the substitution and write the result in terms of \theta with no square root.

14. \displaystyle \sqrt{16 - x^2} with x = 4\sin\theta, \displaystyle -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}
15. \displaystyle \sqrt{4x^2 + 49} with \displaystyle x = \frac{7}{2}\tan\theta, \displaystyle -\frac{\pi}{2} < \theta < \frac{\pi}{2}
16. \displaystyle \sqrt{x^2 - 36} with x = 6\sec\theta, \displaystyle 0 \le \theta < \frac{\pi}{2}
17. \displaystyle \frac{x^2}{\sqrt{1 - x^2}} with x = \sin\theta, \displaystyle -\frac{\pi}{2} < \theta < \frac{\pi}{2}
18. \displaystyle \frac{\sqrt{4 - 9x^2}}{x} with \displaystyle x = \frac{2}{3}\sin\theta, \displaystyle -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}, \theta \neq 0
19. \displaystyle \sqrt{x^2 - 36} with x = 6\sec\theta, but now for x \le -6, using \displaystyle \frac{\pi}{2} < \theta \le \pi
20. \displaystyle \frac{1}{1 + x^2} with x = \tan\theta, \displaystyle -\frac{\pi}{2} < \theta < \frac{\pi}{2}

Reasoning

21. Explain why the substitution x = 3\sin\theta comes with the interval \displaystyle -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}. What would go wrong with \displaystyle \frac{\pi}{2} \le \theta \le \frac{3\pi}{2}?
22. Emiliano writes \displaystyle \frac{\sin^2 x}{1 - \cos x} = 1 + \cos x and says the two sides are the same function. Where is he right, and where not?
23. Explain why cancelling a common factor can only give a short form that exists at least everywhere the original does, while multiplying top and bottom by a new factor, as in Example 5, can lose inputs.